Question

The function '$g$' defined by $g\left(x\right)=\sin \alpha +\cos \alpha -1$ where $\alpha ={\sin }^{-1}\sqrt{\left\{x\right\}}$ is (where {.} denotes fractional part function)

Answer 1

$\begin{array}{c}g\left(x\right)=\sin \alpha +cos\alpha -1\\ \alpha =si{n}^{-1}\sqrt{\left\{x\right\}}\end{array}$
$\left\{-x\right\}=1-\left\{x\right\}$ from the graph of $\left\{x\right\}$
$\begin{array}{c}0\le \left\{x\right\}<1\\ 0\le \sqrt{\left\{x\right\}}<1\\ 0\le {\sin }^{-1}\sqrt{\left\{x\right\}}<\frac{\pi }{2}\\ g\left(x\right)=\sin \left({\sin }^{-1}\sqrt{\left\{x\right\}}\right)+\cos \left({\sin }^{-1}\sqrt{\left\{x\right\}}\right)-1\\ =\sqrt{\left\{x\right\}}+\sqrt{\left(1-\sqrt{\left\{x\right\}}\right)}-1\\ g\left(-x\right)=\sqrt{\left\{-x\right\}}+\sqrt{\left(1-\sqrt{\left\{-x\right\}}\right)}-1\\ =\sqrt{1-\left\{x\right\}}+\sqrt{\left\{x\right\}}-1\\ g\left(x\right)=g\left(-x\right)\end{array}$
even function
$g\left(x+T\right)=\sqrt{\left\{x+T\right\}}+\sqrt{1-\left\{x+T\right\}}-1$
if $T$ is integer
$\begin{array}{c}g\left(x+T\right)=\sqrt{\left\{x\right\}}+\sqrt{\left(1-\sqrt{\left\{x\right\}}\right)}-1\\ =g\left(x\right)\end{array}$
periodic with integer period
not other wise